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Prove that for every positive integer $n$ there is a positive integer with the following characteristics:

I) It has exactly n digits.

II) None of the digits are zero.

III) It is divisible by the sum of its digits.

Obs.: The base considered is base 10.


What I've been able to do so far is this:

The number $5^m$ has $d=\lfloor m\cdot log_{10}(5)\rfloor+1$ digits. Let $a$ be a positive number with $n$ digits such that

1) $n\geq m+1$

2) The last digits of $n$ are the representation of $5^m$ in the decimal base. For example, 2341625 has as its last digits the representation of $5^4=625$ in the decimal base.

3) The sum of its digits is $5^{d'}$ with $d'\leq d$.

If the number $5^m$ in the decimal base doesn't have $0$ as one of its digits, then $a$ will satisfy the conditions of the problem.

For example: Let $m=3$ and $s(a)$ the sum of the digits of $a$. We have $5^3=125$.

$s(a)\in\{5^3,5^2,5,5^0\}$(see 3)).

$a=b_1\cdots b_k125$ where $b_1,\cdots,b_k$ are the first digits of $a$. None of them are zero and $1\leq b_i\leq 9$, with $1\leq i\leq k$, therefore $k\leq b_1+\cdots+b_k\leq 9k$.

$s(a)=b_1+\cdots+b_k+1+2+5\in\{5^3,5^2,5,5^0\}\Rightarrow b_1+\cdots+b_k\in\{117,17,-3,-7\}\Rightarrow $

$b_1+\cdots+b_k\in\{117,17\}$

If $b_1+\cdots+b_k=117$, then $k\leq 117\leq 9k\Rightarrow 13\leq k\leq 117$.

If $b_1+\cdots+b_k=17$, then $k\leq 17\leq 9k\Rightarrow 2\leq k\leq 17$.

Looking at the two inequalities above and knowing that $a=b_1\cdots b_k125$, we prove the problem for $5\leq n\leq 120$.

To be clearer:

Choose $k=7$, then $2\leq k=7\leq 17$. So is possible find $b_1,\cdots,b_7$ such that $b_1+\cdots+b_7=17$.

Choose $b_1=\cdots=b_6=2$ and $b_7=5$, for example. Then $a=2222225125$ satisfies the conditions of the problem.

One problem I can't work around is when $5^m$ has $0$ as one of its digits. That's why I can't solve the problem.


This problem is found in a Brazilian journal geared toward mathematics olympiads. In the issue in which this problem appears there is the introduction of Klarner's theorem. So maybe there is a solution using this theorem, but I can not figure out how to use it in solving this problem.

Klarner's theorem: Let $a,m,p$ be given positive integers. If we can cover a $m \times n$ board using $1 \times p$ pieces, with no leftovers or overlapping pieces, then $p$ divides $m$ or $p$ divides $n$

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    $\begingroup$ Researching Niven / Harshad numbers should be useful: oeis.org/A005349 $\endgroup$ – Alexis Olson Nov 30 '17 at 7:41
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    $\begingroup$ I discovered that this problem is on the short list of IMO 1998. That's why I was able to find a solution which is found in the following link artofproblemsolving.com/community/c6h18497p124442 $\endgroup$ – user477271 Nov 30 '17 at 16:52

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